Problem: In the right triangle shown, $AC = 6$ and $BC = 4$. What is $AB$ ? $A$ $C$ $B$ $6$ $4$ $?$
Explanation: We know $a^2 + b^2 = c^2$ We want to find $c$ ; let $a = 4$ and $b = 6$ So $c^2 = 4^2 + 6^2 = 52$ Then, $c = \sqrt{52}$ Simplifying the radical gives $c = 2\sqrt{13}.$